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I'm interested in the following problem from Artin's Algebra:

Is $O_2$ isomorphic to the product group $SO_2 \times \{ \pm I\}$?

When the field taken is the real numbers the answer is no, which can be seen by counting elements.

I was wondering what's the answer for a general field $F$. Thanks!

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Remark that $-I_2$ is an element of $SO_2$ (maybe you meant something else though), now assume that (on some field $F$) we have :

$$O_2(F)=SO_2(F)\times \langle M \rangle$$

With $M$ being of determinant $-1$ and of order $2$. Remark that if $F$ is of characteristic $2$ we have $O_2(F)=SO_2(F)$ (because $-1=1$) so there is nothing to prove. From now on the characteristic of $F$ is not $2$. If the direct product decomposition holds then :

$$Z(O_2(F))\text{ is of cardinal } >2 \text{ since it contains } (I_2,I_2),(-I_2,I_2),(I_2,M),(-I_2,M)$$

We will show that $Z(O_2(F))$ is always of cardinal $2$. Indeed take $z\in Z(O_2(F))$ then $z$ commutes with $\begin{pmatrix}1&0\\0&-1\end{pmatrix}$, it follows that $z$ is diagonal (here we use the fact that $1\neq -1$) :

$$z=\begin{pmatrix}\lambda_1&0\\0&\lambda_2\end{pmatrix} $$

Now $z$ commutes with $\begin{pmatrix}0&1\\1&0\end{pmatrix}$ and it follows that $\lambda_1=\lambda_2$ hence $z=\pm I_2$ so that $|Z(O_2(F))|=2$.

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